How do I solve for the specific heat in a heat transfer problem?

[mathzeroh]

how're all the fine scientists feeling today?

well i jst had a question regarding my specific heat & Co. homework.

ok here it is: "When a 25g block of metal alloy at 215 degrees celcius (i don't know how to make that round degree symbol on here) is dropped into 85g of water at 22 deg.C, the final temperature is 37 deg.C. What is the specific heat of the alloy?

i worked it out 2 different times and got 2 different answers each time.
once i got 15 J/(g*degrees Celcius) and then the second time i got 1.2 J/(g*degrees Celcius)

any help at all appreciated! thanks in advance!

 

[dextercioby]

Okay.I don't use sign conventions for given and received heat,so i'll simply write

Q_{\mbox{given}}=m_{\mbox{alloy}}c_{\mbox{alloy}}(215-37)

Q_{\mbox{received}}=m_{\mbox{water}}c_{\mbox{water}}(37-22)

Set the 2 #-s equal and then solve for the unknown.

Daniel.

P.S.Pay attention with the units.I'd advise u to use SI-mKgs.

 

[mathzeroh]

wouldnt it be (37-215) because delta T is always t final minus t initial??

and what did you mean by SI-mKgs ?

 

[dextercioby]

Nope,i don't use convention signs.The heats are always positive and equal...(in this 2 body thermal interraction).

SI-mKgs from Système International-metre,Kilogramme,sécond .

Daniel.

 

[mathzeroh]

oh.

well when heat is negative, it meant that its lost, right? that's what i was thought. let me get another crack at it with what you said. but i don't understand why u would have to set them equal together, since Q of water plus Q of metal alloy should equal 0 right?

 

[Gokul43201]

A positive number can never be equal to a negative number. So, what Dexter is doing is simply making sure they are both positive, and equating them.

If you want to strictly follow convention, you would swap Tf and Ti inside the brackets and then put a minus sign in front of one of the Q's. This will give the same result (the two minus signs will cancel out).